Calculus: Analytic Geometry and Calculus, with Vectors

158 Functions, limits, derivatives

Very much more about derivatives remains to be learned, and we give

a modest but important contribution to theory by proving the following

theorem.

Theorem 3.58 If f(x) exists, then f must be continuous at x.

Our hypothesis and a theorem on limits enable us to write

limo [f (x + Ax)- f (x)l = limo x)

• Ax)Ox

=rii i of (x + AX) - f (x)1[ ln

oAxl =

f' (x) .0 = 0.

Therefore,

lim AX + Ox) = Ax),

az- o

and it follows from this that f is continuous at x.

The attainment of a technique for differentiating accurately and

efficiently is of prime importance in the calculus. When we are called

upon to evaluate the left member of the equation

d x (1 + x2)(1) - x(2x)__ 1 - x2

dx1 +x2 (1 +x2)2 (1

+x2)2'

we should say to ourselves "the derivative with respect to x of u (mean-

ing x) over v (meaning 1 + x2) is equal to the quotient with denominator

v2 [write (1 + x2) 2] and numerator v [write (1 + x2)] times du/dx [write 1]

minus u [write x] times dv/dx [write 2x]." We must learn to talk to our-

selves in such a way that we can quickly produce such results as

d l + x2 x(2x) - (1 + x2) _x2 - 1

dx x x2 x2

d x - (1 - x2)(1) - x(-2x) _ 1 + x2

dx 1 - x2 (1 - x2)2 (1 - x2)2

d (^1) 0-1(2x) 2x
TX 1 + x2 (1 + x2)2 (1 -+X2)2'
With the formula for the derivative of a product in mind, we obtain
dx(x2-x+1)(x2+2x+1) = (x2-x+1)(2x+2)

• (x2 + 2x + 1)(2x - 1)
  by saying "the derivative with respect to x of u (meaning x2-- x + 1)

times v (meaning x2 + 2x + 1) is equal to u (write x2- x + 1) times

dv/dx (write 2x + 2) plus v (write x2 + 2x + 1) times du/dx (write

2x - 1)."